Statistics / Probability and Statistical Inference 9 / Chapter 7.6 / Problem 10E

Probability and Statistical Inference | 9th Edition | ISBN: 9780321923271 | Authors: Robert V. Hogg, Elliot Tanis, Dale Zimmerman

We would like to fit the quadratic curve y = ?1 +?2x +

Chapter 7, Problem 10E

(choose chapter or problem)

We would like to fit the quadratic curve $y=\beta_{1}+\beta_{2} x+\beta_{3} x^{2}$ to a set of points $\left(x_{1}, y_{1}\right),\left(x_{2^{\prime}} y_{2}\right), \ldots,\left(x_{n^{\prime}} y_{n}\right)$
by the method of least squares. To do this, let

$h\left(\beta_{1}, \beta_{2}, \beta_{3}\right)=\sum_{i=1}^{n}\left(y_{i}-\beta_{1}-\beta_{2} x_{i}-\beta_{3} x_{i}^{2}\right)^{2}$

(a) By setting the three first partial derivatives of $h$ with respect to $\beta_{1^{\prime}} \beta_{2}$, and $\beta_{3}$ equal to 0 , show that $\beta_{1^{\prime}} \beta_{2}$, and $\beta_{3}$ satisfy the following set of equations (called normal equations), all of which are sums going from 1 to $n$;

$\beta_{1} n+\beta_{2} \sum x_{i}+\beta_{3} \Sigma x_{i}^{2}=\sum y_{i}$

$\beta_{1} \sum x_{i}+\beta_{2} \sum x_{i}^{2}+\beta_{3} \Sigma x_{i}^{3}=\sum x_{i} y_{i}$

$\beta_{1} \sum x_{i}^{2}+\beta_{2} \sum x_{i}^{3}+\beta_{3} \sum x_{i}^{4}=\sum x_{i}^{2} y_{i}$.

(b) For the data

$(6.91,17.52)(4.32,22.69)(2.38,17.61)(7.98,14.29)$

$(8.26,10.77)(2.00,12.87)(3.10,18.63)(7.69,16.77)$

$(2.21,14.97)(3.42,19.16)(8.18,11.15)(5.39,22.41)$

$(1.19,7.50)(3.21,19.06)(5.47,23.89)(7.35,16.63)$

$(2.32,15.09)(7.54,14.75)(1.27,10.75)(7.33,17.42)$

$(8.41,9.40)(8.72,9.83)(6.09 .22 .33)(5.30,21.37)$

$(7.30,17.36)$

$n=25, \Sigma x_{i}=133.34, \Sigma x_{i}^{2}=867.75, \sum x_{i}^{3}=6197.21$

$\sum x_{i}^{4}=46,318.88, \Sigma y_{i}=404.22, \Sigma x_{i} y_{i}=2138.38$ and $\sum x_{i}^{2} y_{i}=13,380.30$. Show that$a=-1.88, b=9.86$, and $c=-0.995$.

(d) Calculate and plot the residuals. Does linear regression seem to be appropriate?

(e) Show that the least squares quadratic regression line is $\hat{y}=-1.88+9.86 x-0.995 x^{2}$.

(f) Plot the points and this least squares quadratic regression curve on the same graph.

(g) Plot the residuals for quadratic regression and compare this plot with that in part (d).

Equation Transcription:

$y={\beta{}}_{1}+$ ${\beta{}}_{2}x+{\beta{}}_{3}{x}^{2}$

$\left({{x}_{1},{y}_{1}}\right),\left({{x}_{2},{y}_{2}}\right),...,\left({{x}_{n},{y}_{n}}\right)$

$h\left({{\beta{}}_{1},{\beta{}}_{2},{\beta{}}_{3}}\right)=$ $\sum\limits_{i=1}^{n}({y}_{i}-{\beta{}}_{1}-{\beta{}}_{2}{x}_{i}-{\beta{}}_{3}{x}_{i}^{2}{)}^{2}$

${\beta{}}_{1},{\beta{}}_{2}$

${\beta{}}_{3}$

$n$

${\beta{}}_{1}n+{\beta{}}_{2}\sum\limits_{\ }^{\ }\ {x}_{i}+{\beta{}}_{3}\sum\limits_{\ }^{\ }\ {x}_{i}^{2}\ \ =\sum\limits_{\ }^{\ }\ {y}_{i}\ ;$

${\beta{}}_{1}\sum\limits_{\ }^{\ }\ {x}_{i}+{\beta{}}_{2}\sum\limits_{\ }^{\ }\ {x}_{i}^{2}+{\beta{}}_{3}\sum\limits_{\ }^{\ }\ {x}_{i}^{3}\ \ =\sum\limits_{\ }^{\ }\ {x}_{i}{y}_{i};\$

${\beta{}}_{1}\sum\limits_{\ }^{\ }\ {x}_{i}^{2}+{\beta{}}_{2}\sum\limits_{\ }^{\ }\ {x}_{i}^{3}+{\beta{}}_{3}\sum\limits_{\ }^{\ }\ {x}_{i}^{4}\ \ =\sum\limits_{\ }^{\ }\ {x}_{i}^{2}{y}_{i}$ .